Question

Three monodisperse Nylon 6 samples with molar masses 10000 g/mol, 30000 g/mol and 60000 g/mol are mixed in a proportion of 1:1:2 by number of chains. The polydispersity index of the resulting sample (rounded off to 2 decimal places) is ______________________________.

Hint:

For discrete mixtures by number of chains, use $M_n=\frac{\sum n_iM_i}{\sum n_i}$ and $M_w=\frac{\sum n_iM_i^2}{\sum n_iM_i}$, then $\text{PDI}=M_w/M_n$.

Answer 1

Correct Answer: 1.27

Number fractions by chains: $n_1:n_2:n_3=1:1:2$ \ ($\sum n_i=4$).
$M_1=10000,\ M_2=30000,\ M_3=60000\ \text{g/mol}$.
Number-average molar mass:\ $M_n=\dfrac{\sum n_i M_i}{\sum n_i}=\dfrac{(1)(10000)+(1)(30000)+(2)(60000)}{4}=\dfrac{160000}{4}=40000\ \text{g/mol}$.
Weight-average molar mass:\ $M_w=\dfrac{\sum n_i M_i^2}{\sum n_i M_i}=\dfrac{(1)(10000)^2+(1)(30000)^2+(2)(60000)^2}{160000}=\dfrac{8.2\times10^9}{1.6\times10^5}=51250\ \text{g/mol}$.
PDI:\ $=\dfrac{M_w}{M_n}=\dfrac{51250}{40000}=1.28125\Rightarrow \boxed{1.28}$.